F 2.50 × 10−2 g of solid Fe(NO3)3 is added to 100. ML of a 1.0 × 10−4 M NaOH solution, will a precipitate form? (ksp=4 × 10−38 for Fe(OH)3)

A precipitate will be formed

Explanation:

The Ksp equilibrium of Fe(OH)₃ is:

Fe(OH)₃ (s) ⇄ Fe³⁺(aq)+ 3OH⁻(aq)

And its expression is:

Ksp = 4x10⁻³⁸ = [Fe³⁺] [OH⁻]³

Where the concentrations are concentrations in molarity in equilibrium,

We can write Q as:

Q = [Fe³⁺] [OH⁻]³

Where [] are actual concentrations in molarity of each specie.

When Q>= Ksp; a precipitate is formed,

When Q< Ksp no precipitate is produced:

[OH⁻] = [NaOH] = 1.0x10⁻⁴M

[Fe²⁺] = 2.50x10⁻²g * (1mol / 179.85g) / 0.100L = 1.39x10⁻³M

179.85g/mol is molar mass of Fe(NO₃)₂ and the volume of the solution is 0.100L = 100mL

Q = [Fe³⁺] [OH⁻]³

Q = [ 1.39x10⁻³] [ 1.0x10⁻⁴]³

Q = 3.8x10⁻¹⁵

As Q >> Ksp; A precipitate will be formed

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explanation:

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